# What is the relation between a basis transformation and an induced transformation $\psi(\Lambda^{-1} p)$ on the wave function? [closed]

I'm having trouble understanding why is $\psi(\Lambda^{-1}p')$ the correct wave function in the Lorentz transformed frame $p' = \Lambda p$.

Suppose the state in frame $O$ is given by

$$|\Psi\rangle = \int dp\, \psi(p)\, |p\rangle$$

then in frame $O'$ with $p' = \Lambda p$, the same state has representation

$$|\Psi\rangle = \int dp'\, \psi'(p')\, |p'\rangle$$

where $|p'\rangle = U(\Lambda) |p\rangle =|\Lambda p\rangle$ and $\psi'(q') = \langle q' | \Psi \rangle$ (all quantities unprimed $p, q, . . .$ are in frame $O$, all quantities primed $p', q', . . .$ are in frame $O'$). Calculating now

$$\langle q' | \Psi \rangle = \langle q | U^{\dagger}(\Lambda)|\Psi\rangle = \int dp\, \psi(p)\, \langle q | U^{\dagger}(\Lambda)\, |p\rangle = \int dp\, \psi(p)\, \delta(q - \Lambda^{-1} p) = \psi(\Lambda q)$$

where the last step follows since $q - \Lambda^{-1} p = 0$ gives $p = \Lambda q$. So all in all $\psi'(q') = \psi(\Lambda q)$ but instead we know the correct answer is $\psi'(q') = \psi(\Lambda^{-1} q')$. Where is the mistake in the reasoning?

Note: similar questions have been asked before but none addresses the issue I'm raising, namely, how to convince oneself that the function space transformation $\psi(\Lambda^{-1} q')$ arises in the way shown above.

• Related (possible duplicate?): physics.stackexchange.com/q/95837 Commented Feb 16, 2015 at 14:15
• Thanks, it's related (as are many others as I noted) but I'm afraid it does not address the issue I raised. I am interested in seeing in detail how the said transformation on a wave function arises due to a passive basis transformation $|p \rangle \mapsto |p' \rangle$ in the Hilbert space, i.e. as a result of calculating the representation $\langle p' | \Psi \rangle$ of the wave function in the new basis. Commented Feb 17, 2015 at 3:03
• You seem to mix something up here - if this is QFT, there is no wavefunction, and also, no momentum basis in the way you wrote it, since the Hilbert space of QFT is always the entire Fock space. Commented Feb 18, 2015 at 15:15
• No I am not mixing up---this is relativistic QM, not QFT. Commented Feb 19, 2015 at 2:41

For the mass $m>0$, spin $j$ representation of the (universal cover of the) Poincaré group, the theory of induced representations give the transformation law $$(u(A,a)\phi)(p) = e^{ip\cdot a} D^{(j)}(A_pAA_{\Lambda(A^{-1})p})\phi(\Lambda(A^{-1})p),\qquad \forall\phi\in L^2(\Omega_m^+,\text d\Omega_m^+)\otimes\mathbb C^{2j+1}$$ where $(A,a)\in SL_2(\mathbb C)\ltimes\mathbb R^n$, $D^{(j)}$ is the spin-$j$ irreducible representation of the little group $SU(2)$ (recall that I have assumed $m>0$ here for definiteness), $p\mapsto A_p$ is any section from the orbit $\Omega_m^+$ to Lorentz transformations in $SL_2(\mathbb C)$ (the exact details of these objects come from the Mackey-Wigner theory of induced representations) and $\text d\Omega_m^+$ is the invariant measure.
For spin $j=0$, this transformation law reduces to $$(u(A,a)\phi)(p) = e^{ip\cdot a}\phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+,$$ and if we neglect the action of the translation semigroup this further reduces to $$(u(A,0)\phi)(p) = \phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+.$$ Taking a Fourier transform to go from momentum space to position space, the transposed action is seen to be $$(u(A,0)\psi)(x) = \psi(\Lambda(A^{-1})x),\qquad x\in\mathbb R^4,$$ which is the special case of the more general relation $$(u(A,a)\psi)(x) = \psi(\Lambda(A^{-1})(x-a)),\qquad x\in\mathbb R^4.$$
• Thanks, this is informative but I'm afraid does not answer the question: how does the said transformation on the wave function arise in the Hilbert space as a result of a passive basis change? That is, how to arrive at the first equation, and in particular $\phi(\Lambda(A^{-1}) p)$, in the first place? Commented Feb 17, 2015 at 3:16
• Let me try to explain the problem from a different perspective. I am already assuming that we have Wigner's (1939) result, that we know how the boost $U(\Lambda)$ acts on a basis vector in the Hilbert space [1], $$U(\Lambda) |p, \sigma \rangle = \sum_{\sigma'} |\Lambda p, \sigma'\rangle D_{\sigma' \sigma}(W(\Lambda, p))$$ Take a boost with no rotation, D = \mathbb{1}, then the basis vector transforms $$U(\Lambda) |p, \sigma \rangle = |\Lambda p, \sigma\rangle$$ Commented Feb 17, 2015 at 10:07
• Ignore now the spin part since it undergoes an identity transformation in this particular situation. Question: given the state $|\Psi\rangle$ above, how do you show that the transformed wave function is given by $\psi(\Lambda^{-1} p')$ or by your last but one equation? [1] W.-K. Tung (1985). Group Theory in Physics. Commented Feb 17, 2015 at 10:07
• In order to answer to this question we have to use rather formal terms, starting from $\langle p'|p\rangle = \delta(p'-p)$. Given this we then have $\langle p'|\Lambda p\rangle=\delta(p'-\Lambda p)=\delta(\Lambda(\Lambda^{-1}p'-p))=\delta(\Lambda^{-1}p'-p)$, since $\det(\Lambda)=1$. Hence the transformation sends $\psi_p(p')$ to $\psi_p(\Lambda^{-1}p')$. Commented Feb 17, 2015 at 10:30